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2 years ago

A. How far does a 100-newton force have to move to do 1,000 joules

Physics

1 answer:

Aloiza [94]2 years ago

5 0

Work done by a force is given as the product of force and the distance moved by the force.

Work done is the product of force and the distance moved by the the force.

Work done = Force × distance

Thus, distance required by the 100 N force is given as:

Distance = work done/force

Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

Force applied = work done/ distance

Force applied by the hoist = 500/2

Force applied by the hoist = 250 N

Distance moved in one push-up = 25 cm = 0.25 m

Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

Distance moved by the force = 0 m

Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

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Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

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1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

$Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110$

The resulting force is negative, meaning the direction is to the left

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3 years ago

A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s

boyakko [2]

Answer:

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3 years ago

5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in

prisoha [69]

The options are;

a. V2 equals 2V1.

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c. V2 equals V1.

d. V2 equals (V1)/4.

e. V2 equals 4V1.

Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

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Thus;

At input;

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At output;

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So, m'1 = m'2

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Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

ρ1•V1•A1 = (ρ1)/2•V2•A2

Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

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A1 and ρ1 will cancel out to give;

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3 years ago

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<u>Explanation: </u>

When both are in a gravity free space, the weights are zero, as we know that the $\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}$

$\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero$

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But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.

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I think the answer is false

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3 years ago